Question

Show that semi-vertical angle of a cone of maximum volume and given slant height is ${\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$.

Answer 1

Let radius, height and slant height of cone be $r,h$ and $l$ respectively.

Therefore, $r^2+h^2=l^2$

Volume of cone (V)=$\frac{1}{3}\pi r^{2}h$

$\therefore V=\frac{\pi }{3}h(l^2-h^2)\Rightarrow \frac{\pi }{3}[l^{2}h-h^3]$

Differentiating w.r.t to h, we get

$\frac{dV}{dh}=\frac{\pi }{3}(l^2-3h^2)$

When $\frac{dV}{dh}=0\Rightarrow h=\frac{1}{\sqrt{3}}$

$\frac{d^{2}V}{d{h}^2}=-2\pi h=-2\pi \frac{1}{\sqrt{3}}\Rightarrow \frac{-2\pi }{\sqrt{3}}<0$

Therefore at $h=\frac{1}{\sqrt{3}}$, volume is maximum.

$\cos \alpha =\frac{h}{l}=\frac{1}{\sqrt{2}}$

$\therefore \alpha ={\cos }^{-1}\frac{1}{\sqrt{3}}.$